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A 35.0-ml sample of 0.20 m lioh is titrated with 0.25 m hcl. What is the ph of the solution after 23.0 ml of hcl have been added

kiruha [24]

<h3><u>Answer;</u></h3>

pH = 12.33

<h3><u>Explanation;</u></h3>

The equation of reaction is :

LiOH(aq) + HCl(aq) --> LiCl(aq) + H2O(l)

Reactants left after the titrant is added;

Total Moles LiOH;

= 0.035L LiOH × (0.2moles/L)

= 0.007moles of LiOH

Moles of HCl;

= 0.023L HCl × (0.25moles/L)

= 0.00575moles HCl is the limiting reagent

Reacting amount of moles of LiOH;

= 0.0575 moles HCl *(1mole LiOH/1moles HCl)

=0.00575 moles LiOH (reacted)

Moles of LiOH left;

= 0.007moles total - 0.00575moles that react

= .00125 moles of LiOH (left)

LiOH is a strong base, which means that it ionizes completely.

0.00125moles LiOH *(moles/0.058L) = 0.02155M of LiOH

LiOH(aq) --> Li+(aq) + OH-(aq)

[LiOH] = [OH-] = 0.02155 M

pOH = -log[OH-]

pOH = -log(0.02155)

pOH= 1.67

pH = 14 - pOH

pH = 14 - 1.67

pH = 12.33

7 0

2 years ago

HELP PLEASE! <br> Picture for question provided

BartSMP [9]

Answer:

Explanation:

The first one is CrO. The Chromium has the same charge as the oxygen so mol numbers are dropped.

The Second one is CrO2 The two oxygens have a charge of 2(-2) = -4. To balance this, the Chromium must have a charge of +4 Cr(Iv)O2

The third one is can be set up like this

Cr + 3(-2) = 0

Cr - 6 = 0

Cr = 6

Therefore the formula is Cr(vi)O3

The last one is a bit tricky. Follow this carefully. There are 2 Crs and 3Os.

The formula looks like this

2Cr + 3(-2) = 0

2Cr - 6 = 0

2Cr = 6

Cr = 3

The formula is Cr(iii)2 O3

5 0

2 years ago

The first excited vibrational energy level of diatomic chlo- rine (Cl2) is 558 cm^-1 above the ground state. Wave- numbers, the

Usimov [2.4K]

Answer:

The answer is "0.0000190 and 2.7 J".

Explanation:

$\to P_4=\frac{e^{-\beta}(4+\frac{1}{2}) hev}{2vb}\\\\$

$=\frac{e^{-9(1.35)}}{0.278v}\\\\=\frac{e^{-12.5}}{0.278}\\\\=\frac{0.000005285}{0.278}\\\\=0.0000190$

Given:

$h=6.626 \times 10^{-34}\\\\c=3\times 10^{10}\\\\v=558\ cm^{-1}\\\\K=1.38 \times 10^{-23}\\\\ T=298\\\\$

$E=\frac{hcv}{KT}\\\\$

by putting the value into the above formula so, the value is 2.7 J

6 0

3 years ago

f 23.6 mL of 0.200 M NaOH is required to neutralize 10.00 mL of a H3PO4 solution , what is the concentration of the phosphoric a

Eduardwww [97]

Answer:

Explanation:

H3PO4(aq) + 3NaOH(aq) → Na3PO4(aq) + 3H2O(l)

mole of NaOH = 23.6 * 10 ⁻³L * 0.2M

= 0.00472mole

let x be the no of mole of H3PO4 required of 0.00472mole of NaOH

3 mole of NaOH required ------- 1 mole of H3PO4

0.00472mole of NaOH ----------x

cross multiply

3x = 0.0472

x = 0.00157mole

[H3PO4] = mole of H3PO4 / Vol. of H3PO4

= 0.00157mole / (10*10⁻³l)

= 0.157M

<h3>The concentration of unknown phosphoric acid is 0.157M</h3>

7 0

2 years ago

Read 2 more answers

Three kilograms of steam is contained in a horizontal, frictionless piston and the cylinder is heated at a constant pressure of

lakkis [162]

Answer:

Final temperature: 659.8ºC

Expansion work: 3*75=225 kJ

Internal energy change: 275 kJ

Explanation:

First, considering both initial and final states, write the energy balance:

$U_{2}-U_{1}=Q-W$

Q is the only variable known. To determine the work, it is possible to consider the reversible process; the work done on a expansion reversible process may be calculated as:

$dw=Pdv$

The pressure is constant, so: $w=P(v_{2}-v_{1} )=0.5*100*1.5=75\frac{kJ}{kg}$ (There is a multiplication by 100 due to the conversion of bar to kPa)

So, the internal energy change may be calculated from the energy balance (don't forget to multiply by the mass):

$U_{2}-U_{1}=500-(3*75)=275kJ$

On the other hand, due to the low pressure the ideal gas law may be appropriate. The ideal gas law is written for both states:

$P_{1}V_{1}=nRT_{1}$

$P_{2}V_{2}=nRT_{2}\\V_{2}=2.5V_{1}\\P_{2}=P_{1}\\2.5P_{1}V_{1}=nRT_{2}$

Subtracting the first from the second:

$1.5P_{1}V_{1}=nR(T_{2}-T_{1})$

Isolating $T_{2}$ :

$T_{2}=T_{1}+\frac{1.5P_{1}V_{1}}{nR}$

Assuming that it is water steam, n=0.1666 kmol

$V_{1}=\frac{nRT_{1}}{P_{1}}=\frac{8.314*0.1666*373.15}{500} =1.034m^{3}$

$T_{2}=100+\frac{1.5*500*1.034}{0.1666*8.314}=659.76$ ºC

7 0

3 years ago